*Physical quantities*

- In

**, we commonly deal with quantities such as pressure, volume, mass, temperature, current, etc. These quantities are known as physical quantities.**

*physical chemistry*- A physical quantity has two components, namely, numerical value and it's unit and is written as,

- The quantity 5 Newton means the physical quantity Newton and its numerical value 5

Appendix physical chemistry |

- The dimension of length = L, mass = M, and time = T.

*Definition of base physical quantities*

*Meter*

- The meter is the length of the path traveled by light in vacuum during a time interval of 1/299 792 458 of a second.

*Unit and dimension of the area*

Area = length × length

Thus the unit of area = unit of length × unit of length

Unit of area in C.G.S. system = cm × cm

= cm²

Unit of the area in S.I. the system is = m × m

= m²

Dimension of area = L × L

= [L]²

Thus the unit of area = unit of length × unit of length

Unit of area in C.G.S. system = cm × cm

= cm²

Unit of the area in S.I. the system is = m × m

= m²

Dimension of area = L × L

= [L]²

*Unit and dimension of Volume*

Volume (V) = length × length × length.

Thus the unit of volume = length × length × length

Unit of area in C.G.S. system = cm × cm × cm

= cm³

Unit of the area in S.I. the system is = m × m × m

= m³

The dimension of area = L × L × L

= [L³]

Thus the unit of volume = length × length × length

Unit of area in C.G.S. system = cm × cm × cm

= cm³

Unit of the area in S.I. the system is = m × m × m

= m³

The dimension of area = L × L × L

= [L³]

*Unit and dimension of density*

From the definition of density (d) = mass/volume

Thus the unit of density = unit of mass/unit of volume

Unit of area in C.G.S. system = gm/cm³

= gm cm⁻³

Unit of the area in S.I. system is = Kg/m³

= Kg m⁻³

Dimension of area = [M]/[L³]

= [M L⁻³]

Thus the unit of density = unit of mass/unit of volume

Unit of area in C.G.S. system = gm/cm³

= gm cm⁻³

Unit of the area in S.I. system is = Kg/m³

= Kg m⁻³

Dimension of area = [M]/[L³]

= [M L⁻³]

*Unit and dimension of force*

From the Newton second law of motion,

force = mass × acceleration

or, force = mass × velocity/time

or, force = mass × (length/time)/time

Here velocity = length/time

∴ Force = mass × (length/time²)

Thus the unit of force in C.G.S. system,

= unit of mass × (unit of length/unit of time²)

= gm × cm/sec² = gm cm/sec² or, gm cm sec⁻²

or, Dyne

Again in the same way unit of force in S.I system,

= Kg m/sec²

or, Kg m sec⁻²

or, Newton

force = mass × acceleration

or, force = mass × velocity/time

or, force = mass × (length/time)/time

Here velocity = length/time

∴ Force = mass × (length/time²)

Thus the unit of force in C.G.S. system,

= unit of mass × (unit of length/unit of time²)

= gm × cm/sec² = gm cm/sec² or, gm cm sec⁻²

or, Dyne

Again in the same way unit of force in S.I system,

= Kg m/sec²

or, Kg m sec⁻²

or, Newton

- From the above discussion dimension of force, = dimension of mass × dimension of length/dimension of time².

Thus the dimension of force

= [M L T⁻²]

= [M L T⁻²]

*Problem*- How to convert 1 newton into a dyne?

1 Newton = Kg m/sec²

= (1 Kg× 1 m )/(1 sec)²

We know that 1 Kg = 10³ gm and 1 m = 10² cm

Thus, 1 Newton = (10³gm × 10² cm)/(1 sec)²

= 10⁵ gm cm/sec2

= 10⁵ Dyne

= (1 Kg× 1 m )/(1 sec)²

We know that 1 Kg = 10³ gm and 1 m = 10² cm

Thus, 1 Newton = (10³gm × 10² cm)/(1 sec)²

= 10⁵ gm cm/sec2

= 10⁵ Dyne

*Unit and dimension of work*

From the definition of work (W)

= force(F) × displacement(d).

Thus, the unit work = unit of force × unit of displacement(d)

∴ Unit of work in C.G.S. system, = (gm cm/sec²) × cm

or, gm cm²/sec²

or, gm cm²sec⁻²

or, erg

And the unit of work in S.I. system = (Kg m/sec²) × m

or, Kg m²/sec²

or, Kg m²sec⁻²

or, Joule

= force(F) × displacement(d).

Thus, the unit work = unit of force × unit of displacement(d)

∴ Unit of work in C.G.S. system, = (gm cm/sec²) × cm

or, gm cm²/sec²

or, gm cm²sec⁻²

or, erg

And the unit of work in S.I. system = (Kg m/sec²) × m

or, Kg m²/sec²

or, Kg m²sec⁻²

or, Joule

- From the above discussion dimension of work, = (dimension of mass × dimension of length)²)/(dimension of time²)

Thus the dimension of work

= [ML²T⁻²]

= [ML²T⁻²]

*Problem*- How to convert 1 joule into erg?

1 Joule = 1 Kg m²/sec²

= (1 Kg × 1 m² )/1 sec²

We know that 1 Kg = 10³ gm and 1 m = 10² cm

Thus, 1 Joule = {10³ gm × (10²)² cm}/1 sec²

= 10⁷ gram cm²/sec²

= 10⁷ erg

= (1 Kg × 1 m² )/1 sec²

We know that 1 Kg = 10³ gm and 1 m = 10² cm

Thus, 1 Joule = {10³ gm × (10²)² cm}/1 sec²

= 10⁷ gram cm²/sec²

= 10⁷ erg

*Unit and dimension of Force*

- The ability to doing work is termed as energy. Thus, the unit and dimension of energy and work are the same.

- The unit of energy in the C.G.S system is erg and S.I. system is Joule

*Unit and dimension of pressure*

From the definition of pressure (P) = force/area

Thus the unit of pressure = unit of force/Unit of area.

Unit of pressure in C.G.S. System = erg/cm²

= erg cm⁻²

Unit of pressure in S.I. system is = joule/m²

= joule m⁻²

Dimension of pressure

= [M L⁻²]

Thus the unit of pressure = unit of force/Unit of area.

Unit of pressure in C.G.S. System = erg/cm²

= erg cm⁻²

Unit of pressure in S.I. system is = joule/m²

= joule m⁻²

Dimension of pressure

= [M L⁻²]